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Distributive Property of Matrices

Let $A$ be an $m\times n$ matrix .  Let $B$ and $C$ be $n\times r$ matrices.

The Distributive Property of Matrices states:

$A\left(B+C\right)=AB+AC$

Also, if $A$ be an $m\times n$ matrix and $B$ and $C$ be $n\times m$ matrices, then

$\left(B+C\right)A=BA+CA$

   Find $A\left(B+C\right)$ :       Find $AB+AC$ :

   $\begin{array}{ccccc}\hfill \begin{array}{r}\hfill \left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\left(\left[\begin{array}{cc}\hfill 0& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]+\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\right)\\ \hfill =\left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\left[\begin{array}{cc}\hfill -2& \hfill -1\\ \hfill 1& \hfill 2\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill 0& \hfill 3\\ \hfill -1& \hfill -2\end{array}\right]\end{array}& \hfill & \hfill & \hfill & \hfill \begin{array}{r}\hfill \left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\left[\begin{array}{cc}\hfill 0& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]+\left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill 2& \hfill 1\\ \hfill -1& \hfill -1\end{array}\right]+\left[\begin{array}{cc}\hfill -2& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill 0& \hfill 3\\ \hfill -1& \hfill -2\end{array}\right]\end{array}\end{array}$

   Find $\left(B+C\right)A$ :       Find $BA+CA$ :

   $\begin{array}{ccccc}\hfill \begin{array}{r}\hfill \left(\left[\begin{array}{cc}\hfill 0& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]+\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\right)\left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill -2& \hfill -1\\ \hfill 1& \hfill 2\end{array}\right]\left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill -2& \hfill -3\\ \hfill 1& \hfill -2\end{array}\right]\end{array}& \hfill & \hfill & \hfill & \hfill \begin{array}{r}\hfill \left[\begin{array}{cc}\hfill 0& \hfill -1\\ \hfill 1& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]+\left[\begin{array}{cc}\hfill -2& \hfill 0\\ \hfill 0& \hfill 1\end{array}\right]\left[\begin{array}{cc}\hfill 1& \hfill 2\\ \hfill 0& \hfill -1\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill 0& \hfill 1\\ \hfill 1& \hfill 1\end{array}\right]+\left[\begin{array}{cc}\hfill -2& \hfill -4\\ \hfill 0& \hfill -1\end{array}\right]\\ \hfill =\left[\begin{array}{cc}\hfill -2& \hfill -3\\ \hfill 1& \hfill 0\end{array}\right]\end{array}\end{array}$

Therefore, $A\left(B+C\right)=AB+AC$ and $\left(B+C\right)A=BA+CA$

Important :   Notice that $A\left(B+C\right)\ne \left(B+C\right)A$ and $AB+AC\ne BA+CA$ .

This is because multiplication of matrices is not commutative. 

The order in which you multiply is important.